Chapter 6.8, Problem 11E

(a).

To determine

To Calculate: The probability that the value of a random variable with an exponential probability density function is more than a certain number.

Answer to Problem 11E

The probability that a customer has to wait for more than $4$ minutes is $0.2$

Explanation of Solution

Given Information: The average waiting time for a customer is $2.5$ minutes

Concept Used:

• If the mean of a random variable $\left(X\right)$ with an exponential probability density function $f\left(x\right)$ is $\text{μ}$ , then
$f\left(x\right)=\frac{1}{u}{e}^{-\frac{x}{u}}$ if $x\ge 0$ and

  $f\left(x\right)=0$ if $x<0$ .

• If $f\left(x\right)$ is a probability density function of a random variable $X$ , then the probability that $X$ lies between $a$ and $b$ , where $a\le b$, is given by
$P\left(a\le X\le b\right)=\underset{a}{\overset{b}{{\displaystyle \int }}}f\left(x\right)dx$

Calculation:

The mean is $\text{μ}=2.5$

So, $f\left(t\right)$ is given by

  $f\left(t\right)=\frac{1}{2.5}{e}^{-\frac{t}{2.5}}$ if $t\ge 0$ and

  $f\left(t\right)=0$ if $t<0$ .

To find the probability that a customer has to wait for more than $4$ minutes, put $a=4$ and b $=\infty$ in the integral for probability to write

  $P\left(4\le T\le \infty \right)=\underset{4}{\overset{\infty }{{\displaystyle \int }}}\frac{1}{2.5}{e}^{-\frac{t}{2.5}}dt$

  $=\frac{1}{2.5}\times {\frac{e^{-\frac{t}{2.5}}}{\left(\frac{-1}{2.5}\right)}]}_4^{\infty }$

  $=-\left(0-e^{-\frac{4}{2.5}}\right)$

  $=0.2$

Therefore, the probability that a customer has to wait more than $4$ minutes is $0.2$

(b).

To determine

To Calculate: The probability that the value of a random variable with an exponential probability density function is less than a certain number.

Answer to Problem 11E

The probability that a customer is served in less than $2$ minutes is $0.55$

Explanation of Solution

Given Information: The average waiting time for a customer is $2.5$ minutes

Concept Used:

• If $f\left(x\right)$ is a probability density function of a random variable $X$ , then the probability that $X$ lies between $a$ and $b$ , where $a\le b$, is given by
$P\left(a\le X\le b\right)=\underset{a}{\overset{b}{{\displaystyle \int }}}f\left(x\right)dx$

Calculation:

From the previous sub-part, $f\left(t\right)$ is given by

  $f\left(t\right)=\frac{1}{2.5}{e}^{-\frac{t}{2.5}}$ if $t\ge 0$ and

  $f\left(t\right)=0$ if $t<0$ .

To find the probability that a customer is served within the first $2$ minutes, put $a=-\infty$ and b $=2$ in the integral for probability to write

  $P\left(-\infty <T\le 2\right)=\underset{0}{\overset{2}{{\displaystyle \int }}}\frac{1}{2.5}{e}^{-\frac{t}{2.5}}dt$

  $=\frac{1}{2.5}\times {\frac{e^{-\frac{t}{2.5}}}{\left(\frac{-1}{2.5}\right)}]}_0^2$

  $=-\left(e^{-\frac{2}{2.5}}-1\right)$

  $=0.55$

Therefore, the probability that a customer is served in less than $2$ minutes is $0.55$

(c).

To determine

To Calculate: The value of a random variable, with exponential probability density function, such that the probability of the random variable being more than the said value is only $2\%$ .

Answer to Problem 11E

The advertisement can say that whoever isn’t served with $9.78$ minutes would get a free hamburger.

Explanation of Solution

Given Information: The average waiting time for a customer is $2.5$ minutes

Concept Used:

• If $f\left(x\right)$ is a probability density function of a random variable $X$ , then the probability that $X$ lies between $a$ and $b$ , where $a\le b$, is given by
$P\left(a\le X\le b\right)=\underset{a}{\overset{b}{{\displaystyle \int }}}f\left(x\right)dx$

Calculation:

From the previous sub-part, $f\left(t\right)$ is given by

  $f\left(t\right)=\frac{1}{2.5}{e}^{-\frac{t}{2.5}}$ if $t\ge 0$ and

  $f\left(t\right)=0$ if $t<0$ .

If $T$ is the time such that the probability of the waiting time being more than $T$ minutes is $2\%$ , then put $a=T$ and $b=\infty$ in the integral for probability to write

  $P\left(2\le T<\infty \right)=\underset{T}{\overset{\infty }{{\displaystyle \int }}}\frac{1}{2.5}{e}^{-\frac{t}{2.5}}dt=2\%$

  $\Rightarrow \frac{1}{2.5}\times {\frac{e^{-\frac{t}{2.5}}}{\left(\frac{-1}{2.5}\right)}]}_T^T=0.02$

  $\Rightarrow -\left(0-e^{-\frac{T}{2.5}}\right)=0.02$

  $\Rightarrow T=-2.5\ln 0.02=9.78$

Therefore, to not have to give free hamburgers to more than $2\%$ of her customers, the advertisement may say that whoever isn’t served with $9.78$ minutes would get a free hamburger.